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Let $X_1, X_2, \ldots, X_n$ be i.i.d. with a common density function:

$$f(x;\theta)=\theta x^{\theta-1}$$ for $0<x<1$ and $\theta>0$. So this is $\operatorname{BETA}(\theta,1)$ distribution.

Given this Maximum Likelihood Estimator (MLE) for $\theta$: $$\hat \theta=\frac{-n}{\sum_{i=1}^n \ln(X_i)}$$

Determine if $\hat\theta$ is biased. If it is biased, could you redefine it to make it unbiased?

Unfortunately, this is where I get stuck; I have no idea how to evaluate the expectation of $\hat\theta$. $$\operatorname E\left( \frac{-n}{\sum_{i=1}^n \ln(X_i)} \right)$$

How does one calculate this?

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  • $\begingroup$ Very closely related to this recent Question. Not voting to say it's a duplicate, but you might get something useful from the Comments there. Short version: $\hat \theta$ is biased. $\endgroup$ – BruceET Feb 15 '18 at 3:03
  • $\begingroup$ I suspected that, but the issue is that I'm not really sure how to mathematically show that it is. $\endgroup$ – jippyjoe333 Feb 15 '18 at 3:06
  • $\begingroup$ Nor am I. I have used simulation to convince myself that neither MLE or MOM is unbiased. Wikipedia is not much help on unbiasedness. Haven't looked at Bayes'. $\endgroup$ – BruceET Feb 15 '18 at 3:22
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    $\begingroup$ It is similar to this math.stackexchange.com/questions/2619937/… and the solution is almost the same! $\endgroup$ – Maffred Feb 15 '18 at 3:46
  • $\begingroup$ If you have any question on the question I linked to I can expand here. $\endgroup$ – Maffred Feb 15 '18 at 3:58
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Define, $$ Y = -\ln X, $$ thus $$ F_Y(y) = \mathbb{P}(Y\le y)=\mathbb{P}(X\ge e^{-y}) = 1-F_X(e^{-y}), $$ hence, $$ f_X(x) = f_X(e^{-y})e^{-y}=\theta e^{-y (\theta - 1)}e^{-y} = \theta e^{-y\theta}, \quad y>0. $$ Therefore, $$ - \sum_{i=1}^n\ln X_i \sim \mathcal{G}amma(n,\theta). $$ As such, $$ \frac{n}{-\sum_{i=1}^n \ln X_i} \sim n\times \mathcal{I}nv\mathcal{G}amma(n, \theta), $$ then the expectation is $$ \mathbb{E}[\hat{\theta}_n] = \frac{n}{n-1}\theta > \theta, $$ namely, $$ \tilde{\theta}_n = \frac{n-1}{n}\hat{\theta}_n, $$ is an unbiased estimator of $\theta$.

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